Date | November 2016 | Marks available | 3 | Reference code | 16N.2.SL.TZ0.T_5 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Calculate | Question number | T_5 | Adapted from | N/A |
Question
A farmer owns a plot of land in the shape of a quadrilateral ABCD.
AB=105 m, BC=95 m, CD=40 m, DA=70 m and angle DCB=90∘.
The farmer wants to divide the land into two equal areas. He builds a fence in a straight line from point B to point P on AD, so that the area of PAB is equal to the area of PBCD.
Calculate
the length of BD;
the size of angle DAB;
the area of triangle ABD;
the area of quadrilateral ABCD;
the length of AP;
the length of the fence, BP.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(BD=) √952+402 (M1)
Note: Award (M1) for correct substitution into Pythagoras’ theorem.
=103 (m) (103.077…, 25√17) (A1)(G2)
[2 marks]
cosBˆAD=1052+702−(103.077…)22×105×70 (M1)(A1)(ft)
Note: Award (M1) for substitution into cosine rule, (A1)(ft) for their correct substitutions. Follow through from part (a).
(BˆAD)=68.9∘ (68.8663…) (A1)(ft)(G2)
Note: If their 103 used, the answer is 68.7995…
[3 marks]
(Area of ABD=)12×105×70×sin(68.8663…) (M1)(A1)(ft)
Notes: Award (M1) for substitution into the trig form of the area of a triangle formula.
Award (A1)(ft) for their correct substitutions.
Follow through from part (b).
If 68.8° is used the area =3426.28… m2.
=3430 m2 (3427.82…) (A1)(ft)(G2)
[3 marks]
area of ABCD=12×40×95+3427.82… (M1)
Note: Award (M1) for correctly substituted area of triangle formula added to their answer to part (c).
=5330 m2 (5327.83…) (A1)(ft)(G2)
[2 marks]
12×105×AP×sin(68.8663…)=0.5×5327.82… (M1)(M1)
Notes: Award (M1) for the correct substitution into triangle formula.
Award (M1) for equating their triangle area to half their part (d).
(AP=) 54.4 (m) (54.4000…) (A1)(ft)(G2)
Notes: Follow through from parts (b) and (d).
[3 marks]
BP2=1052+(54.4000…)2−2×105×(54.4000…)×cos(68.8663…) (M1)(A1)(ft)
Notes: Award (M1) for substituted cosine rule formula.
Award (A1)(ft) for their correct substitutions. Accept the exact fraction 53147 in place of cos(68.8663…).
Follow through from parts (b) and (e).
(BP=) 99.3 (m) (99.3252…) (A1)(ft)(G2)
Notes: If 54.4 and cos(68.9) are used the answer is 99.3567…
[3 marks]